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Core Connections Geometry, 2013

CC

Core Connections Geometry, 2013 View details

1. Section 10.1

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Exercise 21 Page 590

Practice makes perfect

a Let's start by graphing the three coordinates.

We can find the midpoint of the segment by using the Midpoint Formula.

M( x_1+x_2/2, y_1+y_2/2 )

SubstitutePoints

Substitute ( - 1,- 1) & ( 1,9)

M( - 1+ 1/2, - 1+ 9/2 )

Add terms

M(0/2,8/2 )

Calculate quotient

M(0,4)

The midpoint of AB lies at (0,4). Let's also find the midpoint of BC.

M( x_1+x_2/2, y_1+y_2/2 )

SubstitutePoints

Substitute ( 1,9) & ( 7,5)

M( 1+ 7/2, 9+ 5/2 )

Add terms

M(8/2,14/2 )

Calculate quotient

M(4,7)

The midpoint of AB lies at (4,7). Finally, we will add the midpoints D and E to the diagram.

b To calculate the distance between D and E we can use the Distance Formula.

d = sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2)

SubstitutePoints

Substitute ( 4,7) & ( 0,4)

d = sqrt(( 4 - 0)^2 + ( 7 - 4)^2)

▼

Simplify right-hand side

Subtract terms

d = sqrt(4^2 + 3^2)

Calculate power

d = sqrt(16 + 9)

Add terms

d = sqrt(25)

Calculate root

d = 5

The distance of DE is 5 units.

To predict the length of AC, we notice that △ ABC and △ DBE are similar triangles. This is because they have three pairs of congruent angles. The top angle, ∠ B, is shared by the triangles. According to the Reflexive Property of Congruence we know this angle is congruent in our triangles.

Notice that DE ∥ AC. This means ∠ D ≅ ∠ A and ∠ E ≅ ∠ C are congruent by the Corresponding Angles Theorem.

Since DE is the midsegment of AB and BC, it divides these sides in two equal halves. This must mean the sides of △ DBE are half that of △ ABC. Therefore, AC should be twice that of DE. 2DE ⇒ 2( 5)=10 units.

c Like in Part B, we have to use the Distance Formula.

d = sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2)

SubstitutePoints

Substitute ( 7,5) & ( - 1,- 1)

d = sqrt(( 7 -( - 1))^2 + ( 5 - ( - 1))^2)

▼

Simplify right-hand side

a-(- b)=a+b

d = sqrt((7 + 1)^2 + (5 + 1)^2)

Add terms

d = sqrt(8^2 + 6^2)

Calculate power

d = sqrt(64+36)

Add terms

d = sqrt(100)

Calculate root

d = 10

The distance of AC was in fact 10 units.

Subchapter links

10.1.1 Introduction to Chords p.584-585

10.1.2 Angles and Arcs p.589-591

10.1.3 Chords and Angles p.596-597

10.1.4 Tangents and Secants p.601-602

10.1.5 Problem Solving with Circles p.605-606